Proof of Theorem BNS

Notice first that the vectors $\vect{z}_j$, $1\leq j\leq n-r$ are exactly the same as the $n-r$ vectors defined in Theorem SSNS. Also, the hypotheses of Theorem SSNS are the same as the hypotheses of the theorem we are currently proving. So it is then simply the conclusion of Theorem SSNS that tells us that $\nsp{A}=\spn{S}$. That was the easy half, but the second part is not much harder. What is new here is the claim that $S$ is a linearly independent set.

To prove the linear independence of a set, we need to start with a relation of linear dependence and somehow conclude that the scalars involved must all be zero, i.e. that the relation of linear dependence only happens in the trivial fashion. So to establish the linear independence of $S$, we start with \begin{equation*} \lincombo{\alpha}{z}{n-r}=\zerovector. \end{equation*}

For each $j$, $1\leq j\leq n-r$, consider the equality of the individual entries of the vectors on both sides of this equality in position $f_j$, \begin{align*} 0 &=\vectorentry{\zerovector}{f_j}\\ &=\vectorentry{\lincombo{\alpha}{z}{n-r}}{f_j}&&\knowl{./knowls/definition.CVE.knowl}{\text{Definition CVE}}\\ &= \vectorentry{\alpha_1\vect{z}_1}{f_j}+ \vectorentry{\alpha_2\vect{z}_2}{f_j}+ \vectorentry{\alpha_3\vect{z}_3}{f_j}+ \cdots+ \vectorentry{\alpha_{n-r}\vect{z}_{n-r}}{f_j}&&\knowl{./knowls/definition.CVA.knowl}{\text{Definition CVA}}\\ &= \alpha_1\vectorentry{\vect{z}_1}{f_j}+ \alpha_2\vectorentry{\vect{z}_2}{f_j}+ \alpha_3\vectorentry{\vect{z}_3}{f_j}+ \cdots+\\ &\quad\quad \alpha_{j-1}\vectorentry{\vect{z}_{j-1}}{f_j}+ \alpha_{j}\vectorentry{\vect{z}_j}{f_j}+ \alpha_{j+1}\vectorentry{\vect{z}_{j+1}}{f_j}+ \cdots+\\ &\quad\quad \alpha_{n-r}\vectorentry{\vect{z}_{n-r}}{f_j}&&\knowl{./knowls/definition.CVSM.knowl}{\text{Definition CVSM}}\\ &=\alpha_1(0)+ \alpha_2(0)+ \alpha_3(0)+ \cdots+\\ &\quad\quad \alpha_{j-1}(0)+ \alpha_{j}(1)+ \alpha_{j+1}(0)+ \cdots+\alpha_{n-r}(0)&&\text{Definition of $\vect{z}_j$}\\ &=\alpha_{j} \end{align*}

So for all $j$, $1\leq j\leq n-r$, we have $\alpha_j=0$, which is the conclusion that tells us that the only relation of linear dependence on $S=\set{\vectorlist{z}{n-r}}$ is the trivial one. Hence, by Definition LICV the set is linearly independent, as desired.